EDB — 1FG

[1FG] Let \(a=0\) for simplicity. Rewrite the following relations, and prove them.

• If \(n≥ m≥ 1\) then

  \[ O(x^ n)+O(x^ m) = O(x^ m), \quad o(x^ n)+O(x^ m) = O(x^ m),\quad x^ n+O(x^ m) = O(x^ m)\quad . \]

• If \(n{\gt} m≥ 1\) then

  \[ O(x^ n)+o(x^ m) = o(x^ m),\quad x^ n+o(x^ m) = o(x^ m). \]

• For \(n,m≥ 1\)

  \begin{eqnarray*} 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{eqnarray*}
• \[ ∫_ 0^ y O(x^ n)\, {\mathbb {d}}x=O(y^{n+1}) \quad ∫_ 0^ y o(x^ n)\, {\mathbb {d}}x=o(y^{n+1}) \quad . \]

[ [1FH]↺↻]