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We have `10``n`+100 = O(`n`).

*Proof:*
Let c := 110 and m:=1 and take arbitrary `n` >= `m`.
We show

We know

| 10n+100| <=c*|n|.

|10* n+100| = 10*n+100 <= 110n= 110|n| =c|n|.

Author: Wolfgang Schreiner

Last Modification: December 14, 1999