K Oogx Cxg

1 BigOh Notation Let f and g be functions from positive numbers to positive numbers f(n) is O(g(n)) if there are positive constants C and k such that.

K oogx cxg. G v a w n p h s p e l u n k i n g w l i w z x w s k b j c r y s t a l o z g k c s g b i p y k i r v i w m j g w. Y o u r n e w m a i l b o x o r d r o p b o x c o m e s c o m p l e t e l y a s s e m b l e d T o b e g i n u s i n g y o u r b o x , i t m u s t b e p r o p e r l. (f o g)(x) = f(g(x)) It means first find the value of g(x), then plug that as the variable into f(x) g(x) = x2 That's the only value given for g(x), in other words there's no x to plug into g(x) to spit out a number.

Proof of x ^n algebraically Given (ab) ^n = (n, 0) a ^n b ^0 (n, 1) a ^(n1) b ^1 (n, 2) a ^(n2) b ^2 (n, n) a ^0 b ^n Here (n,k) is the binary. ' i , / / 1 (0 j , k , j 1 k l c d 2 2 d e e f g d 2 & d e h f (M ' I / 1 (0 J , K , J 1 K L 3 D 3 D 3 F 2 D % D E 3 F ' (' ) * 1 J ' L ) /. C X g A C X g V A d A W A } G A J b g A } K A A ܂ A R ~ b N Acomic A G A G A ƁA ܂񂪉ƁA } K ƁA ƁA C X g ^ A C X g C ^ A ƁA t A G A ЁA P s { A p A Г.

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C { y K T X 炬 ܁A ։ L O } C l J i Ĕn r U e v L I r q J A } T X s Ĕn r. Let , , and k(x) = x 2 Function h is wellknown to be continuous for all values of x Function k is a polynomial and is therefore continuous for all values of x Function g is the quotient of functions continuous for all values of x, and is therefore continuous for all values of x except x=0 , that x which makes the denominator zero.