2
$\begingroup$

In the expression

$$\int_0^{\infty} \lim_{m \rightarrow \infty} x_m \left( \varepsilon \right) e^{- \varepsilon} \mathrm{d} \varepsilon$$

Is it possible to move the integral inside the Newton's iteration?

$$x_{i + 1} \left( \varepsilon \right) = x_i \left( \varepsilon \right) - \frac{f \left( x_i \left( \varepsilon \right) \right)}{f' \left( x_i \left( \varepsilon \right) \right)}$$

$\endgroup$
1

1 Answer 1

1
$\begingroup$

You need Dominated convergence theorem or monotone convergence theorem for exchanging limits and integrals. That is, either:

  • $x_m(\epsilon)e^{-\epsilon}\le x_{m+1}(\epsilon)e^{-\epsilon} \forall \epsilon\ \forall m$ (or)
  • There should exist an integrable function $f(\epsilon)$ such that $x_m(\epsilon)\le f(\epsilon)\ \forall \epsilon\ \forall m$.
$\endgroup$
2
  • 1
    $\begingroup$ You should say that the function $f(\epsilon)$ must be integrable. $\endgroup$ Commented Feb 13, 2013 at 20:45
  • $\begingroup$ @AntonioVargas: thanks, edited. $\endgroup$
    – Bravo
    Commented Feb 13, 2013 at 20:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .