EDB — 1FG

[1FG] Sia $a=0$ per semplicità. Riscrivete le successive relazioni e dimostratele.

• Se $n\ge m\ge 1$ allora

  $$O(x^n)+O(x^m)=O(x^m),o(x^n)+O(x^m)=O(x^m),x^n+O(x^m)=O(x^m).$$

• Se $n>m\ge 1$ allora

  $$O(x^n)+o(x^m)=o(x^m),x^n+o(x^m)=o(x^m).$$

• Per $n,m\ge 1$

  $$\begin{array}{rcl}{x}^{n}O(x^m)& =& O(x^{n+m})\\ x^{n}o(x^m)& =& o(x^{n+m})\\ O(x^n)O(x^m)& =& O(x^{n+m})\\ o(x^n)O(x^m)& =& o(x^{n+m})\end{array}$$
• $${\int }_0^{y}O(x^n)\mathbb{d}x=O(y^{n+1}){\int }_0^{y}o(x^n)\mathbb{d}x=o(y^{n+1}).$$

[ [1FH]↺↻]